From the top of a tower, 80m high from the ground a stone is thrown in the horizontal direction with a velocity of `8 ms^(1)`. The stone reaches the ground after a time t and falls at a distance of d from the foot of the tower. Assuming `g=10ms^(2)`, the time t and distance d are given respectively by

Options

A. 6s,64m
B. 6s,48m
C. 4s,32m
D. 4s,16m

Explanation

To solve the problem step by step, we will first determine the time taken for the stone to reach the ground and then calculate the horizontal distance it travels during that time.

### Step 1: Determine the time taken to fall (t)

The stone is thrown horizontally from a height of 80 m. The vertical motion of the stone can be described using the equation of motion under gravity:

\[ h = \frac{1}{2} g t^2 \]

Where:
- \( h \) is the height (80 m)
- \( g \) is the acceleration due to gravity (10 m/s²)
- \( t \) is the time in seconds

Rearranging the equation to solve for \( t \):

\[ t^2 = \frac{2h}{g} \]

Substituting the values:

\[ t^2 = \frac{2 \times 80}{10} \]

\[ t^2 = \frac{160}{10} \]

\[ t^2 = 16 \]

Taking the square root of both sides:

\[ t = \sqrt{16} \]

\[ t = 4 \, \text{s} \]

### Step 2: Calculate the horizontal distance (d)

The horizontal distance \( d \) that the stone travels can be calculated using the formula:

\[ d = u \times t \]

Where:
- \( u \) is the initial horizontal velocity (8 m/s)
- \( t \) is the time we just calculated (4 s)

Substituting the values:

\[ d = 8 \times 4 \]

\[ d = 32 \, \text{m} \]

### Final Results

The time taken \( t \) is 4 seconds, and the distance \( d \) is 32 meters.

### Summary

- Time taken \( t = 4 \, \text{s} \)
- Distance \( d = 32 \, \text{m} \)